ON THE CONVERGENCE OF SOME ALTERNATING SERIES
arXiv:1102.4644v1 [math.NT] 23 Feb 2011
ANGEL V. KUMCHEV
1. Introduction This note is motivated by a question a colleague of the author’s often challenges calculus students with: Does the series ∞ X (−1)n | sin n| (1) n n=1
converge? This series combines features of several series commonly studied in calculus: ∞ ∞ ∞ X X X (−1)n | sin n| sin(nx) , and n n n n=1
n=1
n=1
come to mind. However, unlike these familiar examples, the series (1) seems to live on the fringes, just beyond the reach of standard convergence tests like the alternating series test or the tests of Abel and Dirichlet. It is therefore quite natural for an infinite series aficionado to study (1) in hope to find some clever resolution of the question of its convergence. Yet, the author’s colleague reports that although he has posed the above question to many calculus students, he has never received an answer. Furthermore, he confessed that he himself had no answer to that question. As it turns out, there is a good reason for that: the question is quite delicate and is intimately connected to deep facts about Diophantine approximation—facts which the typical second-semester calculus student is unlikely to know. The series (1) is obtained by perturbation of the moduli of the alternating harmonic series, which is the simplest conditionally convergent alternating series one can imagine. In this note, we study the convergence sets of similar perturbations of a wide class of alternating series. In particular, the convergence of (1) follows from our results and classical work by Mahler [6] on the rational approximations to π. Let F denote the class of continuous, decreasing functions f : [1, ∞) → R such that Z ∞ lim f (x) = 0, f (x) dx = ∞. x→∞ 1 P Note that if f ∈ F, then f is a positive function and the alternating series n (−1)n f (n) is conditionally convergent. Our goal is to describe the convergence set of the related series ∞ X (−1)n f (n)| sin(nπα)|. (2) n=1
It is natural to start one’s investigation of (2) with the case when α is rational, since in that case the sequence {(−1)n | sin(nπα)|}∞ n=1 is periodic, and one may hope to see some pattern. Indeed, this turns out to be the case, and one discovers the following result. Theorem 1. Suppose that f ∈ F and that α = a/q, with a ∈ Z, q ∈ Z+ , and gcd(a, q) = 1. The series (2) converges if and only if q is odd. When α is irrational, the convergence of (2) depends on the quality of the rational approximations to α. Thus, before we can state our results concerning irrational α, we need to introduce some terminology. For α ∈ R, let kαk denote the distance from α to the nearest integer, i.e., kαk = min |α − n| : n ∈ Z . 1
Given α < Q, one can construct a unique sequence {an /qn }∞ n=1 of rational numbers such that |qn α−an | = kqn αk and, for all n ≥ 2, min kqαk : 0 < q < qn = kqn−1 αk > kqn αk. The rational numbers an /qn are called best rational approximations to α. The reader can find the detailed construction of the sequence {an /qn }∞ n=1 and some of its basic properties in Cassels [2, §I.2]. In particular, it follows easily from the properties listed in [2] that 1 1 an < α − < . (3) 2qn qn+1 qn qn qn+1 We can now state our main theorem.
Theorem 2. Suppose that f ∈ F and α < Q, and let {an /qn }∞ n=1 be the sequence of best rational approximations to α. Let Qα be the set of even denominators qn such that qn+1 ≥ 2qn . If the series X 1 Z qn+1 f (x) dx (4) q2n 1 qn ∈Qα
converges, then so does the series (2). By combining Theorem 2 with various facts about Diophantine approximation, we obtain the following corollaries. Corollary 3. There is a set D ⊂ R, with Lebesgue measure zero, such that the series (2) converges for all real α < D and all f ∈ F. Corollary 4. Suppose that f ∈ F and α is an algebraic irrationality. Then the series (2) converges. Corollary 5. The series (1) converges. Theorem 2 provides a sufficient condition for convergence of alternating series of the form (2). It is natural to ask how far is this condition from being also necessary. A closer look at the special case f (x) = x−p , 0 < p ≤ 1, reveals that sometimes the convergence of (4) is, in fact, equivalent to the convergence of (2). We have the following result. Theorem 6. Suppose that α < Q, and let {an /qn }∞ n=1 and Qα be as in Theorem 2. When 0 < p ≤ 1, the series ∞ X (−1)n | sin(nπα)|
np
n=1
(5)
converges if and only if the series X 1 Z qn+1 x−p dx q2n 1
(6)
qn ∈Qα
does. In particular, it follows from Theorem 6 that the divergence set of (2) can be uncountable. Indeed, recalling a classical construction used by Liouville [5] to give the first examples of transcendental numbers, we deduce the following corollary. Corollary 7. There is an uncountable set L ⊂ R, dense in R, such that the series (5) diverges for all α ∈ L and all p ∈ (0, 1]. 2
2. Some lemmas from calculus In this section, we collect several technical lemmas needed in the proofs of the theorems. We also need to introduce a couple of pieces of notation. Throughout the remainder of the paper, we write e(θ) = e2πiθ . We also use Landau’s big-O notation: if B > 0, we write A = O(B) if there exists a constant c > 0 such that |A| ≤ cB. In a few places, we will also encounter inequalities like |A| ≤ c(α)B, where c(α) > 0 depends solely on a particular fixed parameter α. In such situations, it is often convenient to slightly abuse the standard terminology and talk of a “constant depending only on α” and to write A = Oα (B). Lemma 1. Suppose that f ∈ F and 1 ≤ X < Y. Then X (−1)n f (n) ≤ f (X). X≤n≤Y
Proof. This follows from the standard proof of the alternating series test. See Bonar and Khoury [1, Theorem 1.75]. Lemma 2. Suppose that f ∈ F, that h, q are integers, with q ≥ 1, and that 1 ≤ X < Y. Then Z X 1 Y f (h + kq) = f (x) dx + O( f (X)). q X X 0 4
p . 1− p
(11)
Proof. Inequality (10) follows from Lemma 7 and the bound Z ν Z ν ν1−p −p −p ≤ t cos t dt t dt = . 1− p 0 0
The closed-form expression for A p is a standard Fourier cosine-transform formula. It can be found in many references on Fourier analysis, though its proof is often omitted. The interested reader will find the most natural proof (which uses the theory of contour integration) in the solution of Problem III.151 in P´olya and Szeg¨o [7, p. 331]. Finally, to derive the lower bound for A p , we use the inequalities sin(πp/2) ≥ p,
Γ(1 − p) =
Γ(1) Γ(2 − p) ≥ . 1− p 1− p
3. Proof of Theorem 1 We derive the theorem from Cauchy’s criterion. Consider the sum S (α; M, N) =
N+M X
(−1)n f (n)| sin(nπα)|,
(12)
n=N+1
where M, N are positive integers. We note that | sin(πan/q)| = | sin(πah/q)| whenever n ≡ h (mod q). Thus, splitting S (a/q; M, N) according to the residue class of n modulo q, we have S (a/q; M, N) =
=
q X
N+M X
(−1)n f (n)| sin(πan/q)|
h=1
n=N+1 n≡h (mod q)
q X
| sin(πah/q)|
N+M X
(−1)n f (n).
(13)
n=N+1 n≡h (mod q)
h=1
In (13), we can express n as n = N + h + kq,
0 ≤ k ≤ K = ⌊(M − h)/q⌋.
Hence, we can rewrite (13) as q K X X Nh (−1)kq f (Nh + kq), (−1) | sin(πah/q)| S (a/q; M, N) =
(14)
k=0
h=1
where Nh = N + h. We now consider separately the cases of even and odd q. Case 1: q odd. Then (−1)kq = (−1)k , and we have K X
(−1)kq f (Nh + kq) =
K X
(−1)k f (Nh + kq).
k=0
k=0
By Lemma 1, the sum on the right side of (15) is bounded by f (N). Thus, it follows from (14) that |S (a/q; M, N)| ≤ q f (N), Since lim f (x) = 0, this establishes the convergence case of Theorem 1. x→∞
5
(15)
Case 2: q even. Then (−1)kq = 1, and we have K K X X f (Nh + kq). (−1)kq f (Nh + kq) =
(16)
k=0
k=0
We apply Lemma 2 to the sum on the right side of (16) and substitute the result into (14) to obtain q If X S (a/q; M, N) = (−1)Nh | sin(πah/q)| + O (q f (N)) , q
(17)
h=1
where I f = I f (M, N) =
Z
N+M
f (x) dx. N
Since q is even and gcd(a, q) = 1, a must be odd. Thus, (−1)h = (−1)ah , and we have q q X X (−1)ah | sin(πah/q)|. (−1)h | sin(πah/q)| =
(18)
h=1
h=1
Note that when x ∈ Z, the expression (−1) x | sin(πx/q)| depends only on the residue class of x modulo q. Also, since gcd(a, q) = 1, the numbers a, 2a, . . . , qa form a complete residue system modulo q (see Hardy and Wright [3, Theorem 56]). Therefore, the sum on the right side of (18) is a rearrangement of the sum q−1 q X X sin(π j(1 + 1/q)). (−1) j | sin(π j/q)| =
(19)
j=1
j=1
An appeal to the well-known formula n X
sin( jθ) =
sin( 21 nθ) sin( 12 (n + 1)θ) sin( 12 θ)
j=1
now yields q−1 X
sin(π j(1 + 1/q)) =
sin((q + 1) π2 ) sin( qπ 2 − sin( π2 +
j=1
=
π 2q )
π 2q )
(−1)q+1 sin(π/2q) = tan (π/2q) . − cos(π/2q)
(20)
Combining (17)–(20), we conclude that when q is even and gcd(a, q) = 1, S (a/q; M, N) =
(−1)N I f tan(π/2q) + O (q f (N)) . q
(21)
To establish the divergence case of Theorem 1, we need to show that there are choices of M and N, with N → ∞, that keep the right side of (21) bounded away from zero. When N is even, (21) yields Z N+M π f (x) dx + O (q f (N)) . S (a/q; M, N) ≥ 2 2q N R∞ Since N f (x) dx diverges, this completes the proof of the theorem. 6
4. Proof of Theorem 2 To prove Theorem 2, we again estimate the sum S (α; M, N) defined by (12). It is convenient to assume that N is even—as we may, since S (α; M, N) = S (α; M, N + 1) + O ( f (N)) . We start by expanding the function | sin(nπα)| in a Fourier series. By Lemma 5, ∞ N+M X 2 X −1 S (α; M, N) = (−1)n f (n)e(αkn). π 4k2 − 1 n=N+1
k=−∞
Using Lemma 1 to estimate the contribution from k = 0 and combining the terms with k = ±m, m ≥ 1, we obtain X ∞ N+M X −1 4 n S (α; M, N) = Re (−1) f (n)e(αkn) + O ( f (N)) . (22) π 4k2 − 1 k=1
n=N+1
We now estimate the contribution to the right side of (22) from terms with k > M. By the triangle inequality and the monotonicity of f , N+M X n (−1) f (n)e(αkn) ≤ M f (N), n=N+1
whence
X M f (N) N+M X M f (N) 1 X n (−1) f (n)e(αkn) = . ≤ 2 2 4k − 1 4k − 1 4M + 2 n=N+1
k>M
k>M
Thus, we deduce from (22) that
4 S (α; M, N) = Re π
X M k=1
M −e(αkN) X n (−1) g(n)e(αkn) + O ( f (N)) , 4k2 − 1
(23)
n=1
where g(x) = f (N + x). By Lemma 4, we have
M−1 M X X ∆g(m)U(β; m), (−1)n g(n)e(βn) = g(M)U(β; M) −
(24)
m=1
n=1
where ∆g(m) = g(m + 1) − g(m) and
U(β; m) =
m m X X e((β + 1/2)n). (−1)n e(βn) = n=1
n=1
Substituting (24) into the right side of (23), we obtain M−1 X 4 S (α; M, N) = Re g(M)V(α; M) − ∆g(m)V(α; m) + O( f (N)), π
(25)
m=1
where
V(α; m) =
M X −e(αkN) k=1
4k2 − 1
U(kα; m).
In order to estimate the right side of (25), we break the sum V(α; m) into blocks depending on the denominators of the rational approximations to α. Let {an /qn }∞ n=1 be the sequence of best rational approximations ∞ that satisfies r to α. We want to extract a subsequence {r j }∞ of {q } n j+1 ≥ 2r j for all j ≥ 1. For every j=1 n=1 n ≥ 1, there is a unique integer k = k(n) ≥ 0 such that 1 ≤ qn+k /qn < 2 ≤ qn+k+1 /qn . 7
(26)
We construct a recursive sequence {n j }∞ j=1 by setting n1 = 1,
n j+1 = n j + k j + 1 ( j ≥ 1),
where k j = k(n j ) is chosen according to (26) with n = n j . If we set r j = qn j , the sequence {r j }∞ j=1 has the desired property. We decompose V(α; m) into blocks V j (α; m) defined by X −e(αkN) V j (α; m) = U(kα; m), (27) 4k2 − 1 k∈K j (M)
where K j (M) is the set of positive integers k subject to k ≤ M and r j−1 < 4k ≤ r j . Next, we obtain three different estimates for V j (α; m). Let q = q(r j ) denote the largest denominator of a best rational approximation to α with q < r j . Note that, by the construction of the r j ’s, we have r j−1 ≤ q < 2r j−1 . Our estimates depend on the size and parity of q. 4.1. Estimation of V j (α; m) for small j. When j is bounded above by an absolute constant, we appeal to (8) and get X kkα + 1/2k−1 = K j (α), say. (28) |V j (α; x)| ≤ 8k2 − 2 r j−1 δa,q (k) − 1 ≥ δa,q (k) . 2 4q 2 X
q/2 0. Since the series (4) converges, we can find an index n0 = n0 (ε) such that Z ∞ X ε 1 qn+1 f (x) dx < . 2 12 n=n0 qn 1 qn ∈Qα
We choose j0 above to be the least integer j ≥ 2 such that r j ≥ qn0 . Then Z Z rj ∞ X X ε 1 qn+1 1 f (x) dx < , f (x) dx ≤ 2 2 q(r j ) 1 12 n=n0 qn 1 j∈Iα (M)
qn ∈Qα 10
(40)
and (40) yields 11ε + Oα,ε ( f (N)). 12 Therefore, we can find an integer N0 = N0 (ε, α, f ) such that when N ≥ N0 , one has |S (α; M, N)|
j0 and b ∈ {1, 2}. Using (25), (28), (31), (32), and (36), we obtain the following version of (38): X S (α; M, N) = (41) S j (α; M, N) + Oα N −p , j∈Iα (M)
where Iα (M) is the set of indices defined in §4.4, M−1 X 4 ′′ ′′ ∆g(m)V j (α; m) , S j (α; M, N) = Re g(M)V j (α; M) − π m=1
V ′′j (α; m)
and is the sum defined by (33). Furthermore, by (34) and the choice of N, for indices j with r j−1 ≤ N, we have ′′ V (α; M, N) ≤ 4q−2 r j ≤ 4q−2 (N + 2), j whence S j (α; M, N) ≤ c5 r−2 N, j−1 for some absolute constant c5 > 0. Thus, from (41), X S (α; M, N) = S j (α; M, N) + Oα (N), (42) j∈I′α (M,N)
where I′α (M, N) is the set of indices j ∈ Iα (M) such that r j−1 > N. We now proceed to obtain an approximation for V ′′j (α; m), which we will then use to estimate the right side of (42). Let a, q, r and θ be as in §4.3. When 2k ≡ q (mod 2q), we have U(kα; m) =
m X
e(knθ) =
m−1 X
e(knθ) + O(1).
n=0
n=1
Thus, using (7) and the Taylor expansion e(z) = 1 + 2πiz + O(|z|2 ), we find that when 2k ≡ q (mod 2q) and k ≤ r, e(kmθ) − 1 e(kmθ) − 1 U(kα; m) = + O(1) = + O(1). 1 − e(kθ) −2πikθ We substitute this approximation in (33) and use Lemma 3 to bound the contribution from the error terms. We obtain X e(kNθ/2) e(kmθ/2) − 1 + O q−2 V ′′j (α; m) = 2 k −1 πikθ k∈L j (M) Z N+m X 1 e(ktθ/2) dt + O q−2 , = 2 k −1 N k∈L j (M)
11
where L j (M) denotes the set of even integers k such that 12 k ∈ K j (M) and k ≡ q (mod 2q). Hence, 4 X Re Ξk (α; M, N) S j (α; M, N) = , π k2 − 1 k∈L j (M)
where Ξk (α; M, N) = g(M)
Z
N+M
e(ktθ/2) dt − N
M−1 X
∆g(m)
m=1
Z
N+m
e(ktθ/2) dt. N
Recall that here g(x) = (N + x)−p . Interchanging the order of summation and integration in Ξk (α; M, N), we find that Z N+M X Ξk (α; M, N) = g(M) − ∆g(m) e(ktθ/2) dt N
=
Z
t−N≤m≤M−1
N+M
g(⌈t⌉ − N)e(ktθ/2) dt. N
Since g(⌈t⌉ − N) − t−p ≤ pt−p−1 by the mean-value theorem, we obtain Z N+M Ξk (α; M, N) = t−p e(ktθ/2) dt + O N −p . N
Hence, after another appeal to Lemma 3 to estimate the contribution from the error terms, we have Z N+M 1 4 X −p −2 −p t cos(πktθ) dt + O q N S j (α; M, N) = π k2 − 1 N k∈L j (M) Z 4 X (πk|θ|) p−1 νk +µk −p = t cos t dt + O q−2 N −p , 2 π k −1 νk
(43)
k∈L j (M)
where νk = πk|θ|N and µk = πk|θ|M. Summing over j, we deduce from (42) and (43) that X (πk|θ j |) p−1 Z νk +µk 4 X t−p cos t dt + Oα (N) , S (α; M, N) = π ′ k2 − 1 νk
(44)
j∈Iα (M,N) k∈L j (M)
where |θ j | = q−1 kqαk, q = q(r j ). In order to estimate the right side of (44), we will impose some restrictions on the choice of M. Let Q′α be the subset of Qα containing those qn for which Z qn+1 x−p dx > q1−p/2 , n Q′′ α
and let convergent
1 ′ = Qα P \ Qα . The contribution to the series (6) from terms with qn ∈ Q′′ α is dominated series q q−1−p/2 . Thus, the divergence of (6) implies the divergence of the series X 1 Z qn+1 x−p dx. 2 q qn ∈Q′α n 1
by the
(45)
In particular, the set Q′α is infinite. We restrict M to the sequence of numbers of the form q1+p/3 , with q ∈ Q′α . ′ Let J = J(M, N) denote the largest index in the set I′α (M, N), and set I′′ α (M, N) = Iα (M, N) \ {J} and q = q(rJ ). Using our restriction on the choice of M, Lemma 3, and the bound Z νk +µk ( −1 1−p if p < 1, −p ≤ (1 − p) µk t cos t dt ln M if p = 1, νk 12
we find that the term with j = J in (44) is bounded above by X q1−p/2 ≤ 2c6 q−1 , c6 k2 − 1 q≤2k≤r J k≡q (mod 2q)
where c6 = c6 (p) > 0 is a constant depending only on p. Hence, X (πk|θ j |) p−1 Z νk +µk 4 X S (α; M, N) = t−p cos t dt + Oα,p (N) . π ′′ k2 − 1 νk
(46)
j∈Iα (M,N) k∈L j (M)
Since the integrals on the right side of (46) behave somewhat differently when p = 1 and when 0 < p < 1, we now consider these two cases separately. 5.1. The case p = 1. When j ∈ I′′ α (M, N) and k ∈ L j (M), we have µk = πk|θ j |M > 1. Hence, by Lemma 7, Z νk +µk Z 1 t−1 cos t dt = t−1 cos t dt + O(1) νk
νk 1
=
Z
t−1 1 + O t2
νk
dt + O(1)
= − ln νk + O(1) = ln r j + O(ln(kN)). From this inequality and (46), we obtain S (α; M, N) = ≥
4 π
X
X
j∈I′′ α (M,N) k∈L j (M)
ln r j + Oα (N) k2 − 1
ln r j + Oα (N) q(r j )2 − 1 j∈I′′ (M,N) α Z X 1 qn+1 dt + Oα (N), q2n 1 t ′
4 π
X
≥
(47)
qn ∈Qα (M,N)
where Q′α (M, N) is the set of those qn ∈ Q′α for which qn > N and qn+1 ≤ M. In view of the divergence of the series (45), this establishes that lim sup S (α; M, N) = ∞. M→∞
5.2. The case 0 < p < 1. When j ∈ I′′ α (M, N) and k ∈ L j (M), by Lemma 8, Z νk +µk 1−p −p t−p cos t dt = A p + O p νk + µk , νk
where A p is the Fourier integral (11). From this inequality and (46), we obtain S (α; M, N) =
4A p π
4A p ≥ 2−p π
X (πk|θ j |) p−1 + Oα,p (N + ∆) k2 − 1
X
j∈I′′ α (M,N) k∈L j (M) 1−p
X
j∈I′′ α (M,N)
rj
q(r j )2
where ∆ = M −p
X
+ Oα,p (N + ∆) ,
X (k|θ j |)−1 . k2 − 1
j∈I′′ α (M,N) k∈L j (M) 13
By (3), Lemma 3, and the restriction on M, ∆ ≤ 4M
X
−p
j∈I′′ α (M,N)
rj ≤ 4M −γ p q(r j )2
1−p
X
j∈I′′ α (M,N)
rj
q(r j )2
,
where γ p = 41 p2 . Thus, for sufficiently large values of M, we obtain S (α; M, N) ≥
2A p π2−p
≥ A′p
r1−p j
X
+ Oα,p (N) , q(r j )2 Z 1 qn+1 −p t dt + Oα,p (N) , q2n 1
j∈I′′ α (M,N)
X
qn ∈Q′α (M,N)
(48)
where A′p = 2π p−2 (1 − p)A p > 0 and Q′α (M, N) is defined as in §5.1. Therefore, once again, using (48) and the divergence of the series (45), we conclude that lim sup S (α; M, N) = ∞. M→∞
This completes the proof of the theorem. 6. Proofs of the corollaries In this section, we derive the corollaries from Theorems 2 and 6. 6.1. Proof of Corollary 3. For δ > 0, let Dδ denote the set of real α such the inequality qn+1 ≤ q1+δ fails n for an infinite number of denominators qn of best rational approximations to α. By a classical theorem on Diophantine approximation due to Khinchin [4], for any fixed δ > 0, the set Dδ has Lebesgue measure zero. Let D = D1/2 . Then, for α ∈ D and f ∈ F, we have Z qn+1 f (x) dx ≤ f (1)qn+1 ≤ f (1)q3/2 n 1
P for all but a finite number of qn ∈ Qα , and the series (4) is dominated by q∈Qα q−1/2 —which converges, because the elements of Qα grow at least exponentially. Therefore, the series (2) converges by Theorem 2.
6.2. Proof of Corollary 4. When α is an algebraic irrationality, by a celebrated result of Roth [8], the inequality qn+1 ≤ q3/2 n holds for all but a finite number of denominators qn of best rational approximations to α. Thus, we can argue as in the proof of Corollary 3. 6.3. Proof of Corollary 5. We need some information about the the best rational approximations to 1/π. By a classical result of Mahler [6], for all a, q ∈ Z with q ≥ 2, |π − a/q| ≥ q−42 .
(49)
If an /qn is a best rational approximation to 1/π, with n sufficiently large, we deduce from (3) and (49) that 1 an 1 q 1 n π − ≥ 1 > 1 , > − > qn qn+1 π qn 10 an 10a42 c7 q42 n
n
where c7 > 0 is an absolute constant. Thus, the denominators of best rational approximations to 1/π satisfy −1 and α = 1/π. In this case, the series (4) takes qn+1 ≤ c7 q41 n . We can now apply Theorem 2 with f (x) = x the form X ln qn+1 . q2n qn ∈Q1/π P By the discussion in the preceding paragraph, this series is dominated by q q−2 ln q, so the convergence of (1) follows from Theorem 2. 14
6.4. Proof of Corollary 7. In 1851, Liouville [5] considered the series ∞ X 10−k! ξ= k=1
and proved that its sum is transcendental, thus furnishing the first known example of a transcendental number. More generally, we let a, q be integers, with gcd(a, q) = 1, and {dk }∞ k=1 be any infinite sequence of 1’s and 3’s, and we consider the series ∞ a X dk 10−k! (m ≥ 1). λ= + q k=m P The partial sums λN = a/q + k≤N dk 10−k! satisfy the inequality |λ − λN | ≤ 3
∞ X
10−k!
0 is a constant depending at most on q. Furthermore, when N is sufficiently large, inequality (50) is possible only if λN = an /qn . Therefore, for large N, the partial sums λN belong to the sequence of best rational approximations to λ; clearly, such λN have even denominators when expressed in lowest terms. This suffices to establish the divergence of the series (6) at α = λ for any fixed p < 1. Therefore, the series (5) diverges at the numbers λ of the above form. Clearly, the set L of all such λ is dense in R, and it is an exercise in elementary set theory to show that L has the cardinality of the continuum. When p = 1, we can use a similar argument, but we need to modify the above construction of the λ’s. In this case, we want the denominators qn to satisfy the inequality ln qn+1 ≥ c8 q2n . One way to achieve that is to replace the factor 10−k! in the definition of λ by 10−bk , where {bk }∞ k=1 is the recursive sequence defined by b1 = 1,
bk+1 = 100bk
(k ≥ 1).
Acknowledgment. The author would like to thank Geoffrey Goodson, Alexei Kolesnikov, Pencho Petrushev, and Houshang Sohrab for several conversations, suggestions and comments during the writing of this paper. References [1] [2] [3] [4] [5] [6] [7] [8]
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